# For what values of $x$ is the series $\sum_{k=2}^{\infty}\frac{(x-2)^k}{k\ln{k}}$ convergent?

For what values of $x$ is the series $$\sum_{k=2}^{\infty}\frac{(x-2)^k}{k\ln{k}}$$ absolutely convergent, conditionally convergent and divergent?

Denote $a_k=(x-2)^k/k\ln{k},$ I get

$$\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_k}\right|=|x-2|\lim_{k\rightarrow\infty}\frac{k\ln{k}}{(k+1)\ln{k+1}}=1,$$

according to L'Hopitals rule. So I have that $-1 \leq x-2\leq1$ which means that the series is absolutely converging for $1<x<3$. Checking the endpoints:

$x=1:\Rightarrow$

$$\sum_{k=2}^{\infty}\frac{(-1)^k}{k\ln{k}}.$$

This is an alternating series. Denote $b_k=\frac{1}{k\ln{k}}.$ I have to show that $\lim_{k\rightarrow\infty}b_k=0$ and that $b_k$ is a decreasing function. The first condition is trivial. The second condition is trivial to since it is easy to see that $b_k>b_{k+1}>b_{k+2}>...>b_{k+m},$ for $m\in\mathbb{N}.$ So this is a conditional convergence.

$x=3:\Rightarrow$

$$\sum_{k=2}^{\infty}\frac{1}{k\ln{k}}$$ and the integraltest gives

$$\int_{x=2}^{\infty}\frac{1}{x\ln{x}}=\lim_{x\rightarrow \infty}\log(\log{x})=\infty$$

which diverges and the series diverges too for $x=3$. Thus the series behaviour can be summed up as follows:

The radius of convergence is $R=1$ and the interval of convergence is $I=[1,3).$ The series is

• Absolutely convergent for $x\in(1,3)$
• Conditionally convergent for $x=1$
• Divergent for the rest.

Is this correct?

EDIT: What I don't understand is that in my book the following theorem about convergence is stated:

• If the series $\sum|a_k|$ converges, then the series $\sum a_k$ also converges. This series is then absolutely convergent.
• If the series $\sum |a_k|$ is divergent and $\sum a_k$ is convergent, then the series $\sum a_k$ is conditionally convergent.

But in my solution above, I don't see anywhere this theorem is applied. I've just gonv for the limit of $|a_{k+1}/a_k|$, not the sum of $|a_k|$. Can someone explain this?

Using ratio test, one claims that for all $x$ with $|x-2|<1$, the series converges absolutely. For $x-2=1$, the series is $\displaystyle\sum_{k=2}\dfrac{1}{k\log k}$ which does not converge. For $x-2=-1$, the series is $\displaystyle\sum_{k=2}\dfrac{(-1)^{k}}{k\log k}$ which is convergent by alternating series test, but not absolutely, so it is conditional.

• Hi, please see my edit above. I'm not sure how to examine the conditional convergence. – Parseval Jan 7 '18 at 2:18
• For the divergent case for $x=3$, you cannot argue by having a divergent series as an upper bound. You may use the integral test, the primitive is $\log(\log x)$. – user284331 Jan 7 '18 at 2:21
• And for $x=1$, it is conditional, because the absolute value of the term would turn to equally $x=3$ case, which is divergent. – user284331 Jan 7 '18 at 2:23
• Hi, can you please check my solution again and also answer the question down in the edit? – Parseval Jan 7 '18 at 10:53
• Of course in dealing with the convergence of power series, it's likely that you use ratio test or the like rather than the definition of conditional or absolute convergent. The conditional or absolutely convergence are specially for endpoints, because by ratio test, all the points lying in the open interval are absolutely, as you may check by the statement of ratio test. For the endpoint $x=1$, actually your argument is not valid, that you didn't check that $\displaystyle\sum_{k=2}\left|\dfrac{(-1)^{k}}{k\log k}\right|$ is divergent, at least in that paragraph you didn't do so. – user284331 Jan 7 '18 at 19:12

Hints:

1. The Bertrand's series $\displaystyle\sum_{k\ge2}\frac1{k\,\ln k}$ is divergent.
2. If $a>1$, $k\ln k=o(a^k)\strut$.

You should say that it's "conditionally convergent for $x=1$" and "absolutely convergent for $x\in \color{red}{(}1,3)$", and I don't believe your argument that $\sum_{k=2}^{\infty}\frac{1}{k\ln{k}}$ diverges is valid.

The comparison that $\frac{1}{k \ln k} < \frac{1}{k}$ would have teeth if $\sum_{k=2}^{\infty}\frac{1}{k}$ converges, and it would show that $\sum_{k=2}^{\infty}\frac{1}{k\ln{k}}$ also converges. As we know it doesn't. If $\frac{1}{k \ln k} > \frac{1}{k}$ were somehow true, then you could argue that since $\sum_{k=2}^{\infty}\frac{1}{k}$ diverges $\sum_{k=2}^{\infty}\frac{1}{k\ln{k}}$ definitely diverges, but we know that's not true either.

The integral test is the most basic way to show it diverges I think. $\bigg [\int \frac{1}{x \ln x} dx=\int \frac{1/x}{\ln x} dx \left(\to \int \frac{du}{u}\right)=\ln(\ln x) \to \infty \ \ \text{as} \ \ x \to \infty \bigg] \implies \bigg [\sum_{k=2}^{\infty}\frac{1}{k\ln{k}} \to \infty \bigg]$

Accordingly, as user284331 pointed out, you should say that $\sum_{k=2}^{\infty}\frac{(x-2)^k}{k\ln{k}}$ converges conditionally at $x=1$

• Hi, can you please check my solution again and also answer the question down in the edit? – Parseval Jan 7 '18 at 10:53
• A first read of what your book says is confusing. As i know it, conditionally convergent means a given series only converges under the condition that the series alternates where otherwise, as with$\sum \frac {1}{k \ln k}$, it would diverge. An absolutely convergent series converges regardless, i.e. whether it alternates or not. – AmateurMathPirate Jan 7 '18 at 11:34
• I see that in your book $a_k$ is alternating already? – AmateurMathPirate Jan 7 '18 at 11:38
• No, not necessarily. This has nothing to do with alternating series yet. – Parseval Jan 7 '18 at 12:07
• @Parseval took a sec to find a series whose terms switch polarity without alternating. Made me consider something like $\sum \frac { \cos \frac {(3\pi k)}{7}}{k}$ . Really never considered those. I assume you're good now with the types of convergence? – AmateurMathPirate Jan 8 '18 at 2:38